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3 years ago

1. La suma de dos números es 115. El número mayor es dos veces más 22 unidades que el otro. Hallar

Mathematics

1 answer:

Lerok [7]3 years ago

6 0

De acuerdo con un sistema de ecuaciones, tiene-se que los números son 31 y 84.

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En el sistema de ecuaciones, tiene-se que los números son x e y.
Suma de 115, o sea, $x + y = 115$
El número mayor es dos veces más 22 unidades que el otro, o sea, $x = 2y + 22$

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Primero se encuenta el número menor, reemplazando la segunda ecuación en la primera:

$x + y = 115$

$2y + 22 + y = 115$

$3y = 93$

$y = \frac{93}{3}$

$y = 31$

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El número mayor es dado en función de el menor, o sea:

$x = 2y + 22 = 2(31) + 22 = 62 + 22 = 84$

Los números son 31 y 84.

Otro problema resuelto por sistema de ecuaciones es dado en brainly.com/question/24637096

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Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$

Observe that, because $n \geq 2$ and is an integer,

$n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5$

In concusion, the statement is true if and only if n is a non negative integer such that $n\leq 77$

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute $(4n)^4- \sum^{n}_{i=0} (2i)^4$ for 77 and 78 you will obtain:

$(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064$

$(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992$

7 0

3 years ago

Okay i need some help fast

Troyanec [42]

Answer:

Dimensions : 1 by 1 by 2 feet,

Volume ⇒ 2 ft^3

Step-by-step explanation:

Here, I can only assume that this rectangular prism is assigned to have dimensions 1 foot by 1 foot by 2 feet;

I can tell that these types of boxes have height and width as the same lengths, so given that;

Dimensions : 1 by 1 by 2 feet,

Volume ⇒ 1 * 1 * 2 ( by definition ) ⇒ 2 ft^3

In other words, Volume ⇒ 2 ft^3

6 0

3 years ago

I need help anyone know

kaheart [24]

Answer:

yes ur correct its nonlinear lol

Step-by-step explanation:

6 0

3 years ago